已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:(1)sinθ-cosθ;(2)sin³(π/2-θ)+cos³(π/2+θ)

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已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)

已知sin(π-θ)-cos(π+θ)=√2/3,其中π/2<θ<π,求下列各式的值:
(1)sinθ-cosθ;
(2)sin³(π/2-θ)+cos³(π/2+θ)
(1)sin(π-θ)-cos(π+θ)=√2/3
sinθ+cosθ=√2/3
(sinθ+cosθ)²=2/9
1+2sinθcosθ=2/9
2sinθcosθ=-7/9
sinθcosθ=-7/18
(sinθ-cosθ)²=1-2sinθcosθ=23/9
因为π/2<θ<π,所以sinθ-cosθ>0,所以sinθ-cosθ=√23/3
(2)sin³(π/2-θ)+cos³(π/2+θ)
=cos³θ-sin³θ
=(sinθ-cosθ)(sin²θ+sinθcosθ+cos²θ)
=√23/3*(1-7/18)
=11√23/54
打得好累啊